   Question Answers

# Find the derivative of $\cos \left( \log x+{{e}^{x}} \right)$.  Hint:For the above question we have been given a composite function to find its derivative and we know that to find the derivative of a composite function we will have to use the chain rule of derivative which states that the derivative of a composite function,
$F\left( g\left( x \right) \right)$ is $F'\left( g\left( x \right) \right)\times g'(x)$

We have been asked to find the derivative of the function, $\cos \left( \log x+{{e}^{x}} \right)$.
Since it is a composite function, so we will use the chain rule of derivative to find its derivatives and the chain rule states that the derivative of a function $F\left( g\left( x \right) \right)$ is equal to $F'\left( g\left( x \right) \right)\times g'(x)$.
So the differentiation of $\cos \left( \log x+{{e}^{x}} \right)$ is shown as follows by applying the chain rule:
$\dfrac{d}{dx}\left[ \cos \left( \log x+{{e}^{x}} \right) \right]=-\sin \left( \log +{{e}^{x}} \right)\times \dfrac{d}{dx}\left( \log x+{{e}^{x}} \right)$
Since we know that the derivative of cos x is –sin x.
$\Rightarrow -\sin \left( \log x+{{e}^{x}} \right)\times \left[ \dfrac{d}{dx}\left( \log x \right)+\dfrac{d}{dx}\left( {{e}^{x}} \right) \right]$
As we know the derivative of log x is $\dfrac{1}{x}$ and the derivative of ${{e}^{x}}$ is ${{e}^{x}}$, we can write the above function as follows.
$\Rightarrow \dfrac{d}{dx}\left[ \cos \left( \log x+{{e}^{x}} \right) \right]=-\sin \left( \log x+{{e}^{x}} \right)\times \left( \dfrac{1}{x}+{{e}^{x}} \right)$
Therefore, the required derivative of the given function is equal to $-\sin \left( \log x+{{e}^{x}} \right)\times \left( \dfrac{1}{x}+{{e}^{x}} \right)$.
Note: Be careful while applying the chain rule to the composite function and also take care of the sign. Sometimes we take the derivative of cos x as sin x by mistake and we just forget the negative sign before sin x, so be careful at that time otherwise we will get incorrect answers.
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